Page 5 - Lesson Note - 1
P. 5
C = A ε o / (d – t + t / K)
If a conducting (metal) slab is inserted between the plates, then
C = A ε o / (d – t)
When more than one dielectric slabs are placed fully between the plates, then
The plates of a parallel plate capacitor attract each other with a force
2
F = Q / 2 A ε o
When 9. dielectric slab is placed between the plates of a capacitor than charge induced on its side
due to polarization of dielectric is
q’ = q (1 – 1 / k)
Capacitors Combination
(i) In Series
Resultant capacitance = 1 / C = 1 / C 1 + 1 / C 2 + 1 / C 3 + ….
In series charge is same on each capacitor, which is equal to the charge supplied by the source.
If V 1, V 2, V 3,…. are potential differences across the plates of the capacitors then total voltage
applied by the
source
V = V 1 + V 2 + V 3 + ….
(ii) In Parallel
Resultant capacitance C = C 1 + C 2 + C 3 + ….
In parallel potential differences across the plates of each capacitor is same.
If q 1, q 2, q 3 are charges on the plate of capacitors connected in parallel then total charge given by
the source
